The solution of $\frac{d y}{d x}+\frac{1}{3} y=1$ is
The solution of $\frac{d y}{d x}+\frac{1}{3} y=1$ is
- $y=3+c e^{x / 3}$
- $y=3+c e^{-x / 3}$
- $3 y=c+e^{x / 3}$
- $y^2+x+x^2+2=c e^{2 x}$
Solution
We have,
$
\begin{array}{rlrl}
& \quad \frac{d y}{d x}+\frac{y}{3} & =1 \\
\Rightarrow & & d y & =-\frac{(y-3)}{3} d x \\
\Rightarrow & \frac{d y}{y-3} & =-\frac{1}{3} d x
\end{array}
$
$\begin{aligned} & \Rightarrow \quad \log (y-3)=-\frac{1}{3} x+\log c \\ & \Rightarrow \quad y-3=c e^{-x / 3} \Rightarrow y=3+c e^{-x / 3}\end{aligned}$
Asked in: AP EAMCET 2002
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